Question

Matlab question:

Resistive networks are well-represented by linear equations. Consider the DC circuit shown below: Figure 1 This problem can be converted into a system of simultaneous linear equations by applying Kirchhoff's law and Ohm's law. In circuit design, i represents current (measured in amperes, A), V represents voltage (in volts, V), and R represents resistance (in ohms, ohm). Kirchhoff's law states that the sum of the currents entering a node is equal to the sum of the currents leaving the node. Applying Kirchoff's law to nodes 1 and 2 respectively yields the equations i_1 = t_2 + i_3 i_2 + i_3 = i_4 To apply Ohm's law, we can use two separate pathways through the circuit in going from the positive side of the DC power source to the negative side: the inner circuit and the outermost circuit. In both cases, the sum of the current times the resistance is equal to the applied voltage V_1. The resulting equations are given as i_1R_1 + i_3R_3 = V_1 i_1R_1 + i_2R_2 = V_1 For this particular system, V_1 has a value of 100 V, R_1 has a value of 10 ohms, R_2 has a value of 5 ohms and R_3 has a value of 10 ohms. The current values are the unknowns in this linear system of simultaneous linear equations. Analytically (i.e. with paper and pencil), put this system into the matrix form Ax = b, and use that result to enter the variables A and b into your program; you do not need to turn in your analytical work for determining what A and b are. Once you have determined what A and b are, solve this system using the matrix inverse. Print the values of the temperatures to the Command Window with labels and units; do not use a loop to automate the printing process.

Answer 1

Explaination:

Given equations are:

i₁ = i₂ + i₃                                            …… (1)

i₂ + i₃ = i₄                                            …… (2)

i₁ R₁ + i₃ R_{3 =} V₁                                          …… (3)

i₁ R₁ + i₂ R_{2 =} V₁                                   …… (4)

Substitute equation (1) in equation (2), the result is:

i₁ = i₄                                                   …… (5)

From equation (1), i₃ equals,

i₃ = i₁ – i₂                                             …… (6)

Given, the value of Vi = 100 V, R,-10 Ω, R,-5 Ω, and R,-10 Ω

Substitute the values in the equation (3)

10 i₁ + 10 i_{3 =} 100

Substitute the equation (6)

                                             

10 ή + 10 (11-12) = 100 10ή + 10h-10i,-100 20 i1-10i,-100

Substitute the values of R₁, R₃, R2, and V1 in the equation (4)

10 i₁ + 5 i_{2 =} 100                                  …… (8)

Convert the equation (7) and (8) in matrix form Ax = b where,

Ax = b form is:

20 -10 10 5 「100 100 Ax = b form is: b= 20-10]「 1100 10 5」L:JL100

In the MATLAB code, Give the values of A and b. Find the inverse of A and multiply by b. Finally print the desired result of current.

Screenshot of the code:

%Substitute the value of A as shown in equation (9). A -[20 -10;10 5]; %substitute the value of b as shown in equation (9). b [100;100]; % solving using inverse 1 = inv(A) *b; %Print the result 11 fprintf('i1-.2fA In',I(1)); %Print the result 12 fprintt('i2 -Xo.2fA \n",1(2)); %print the result 13 = 11-12 as shown in the %equation (6) fprintf('13 .2fA In',I(1)-I (2)); %Print the result 14-11 as shown in the %equation (5) fprintf('i4.2fA\n',I(1));

Sample Output:

sh-4.3$ octave -qf --no-window-system demo.m octave: X11 DISPLAY environment variable not set octave: disabling GUI features warning: function ./demo.m shadows a core library function 12-5.00A 14-7.50A sh-4.3$

Code to copy:

%Substitute the value of A as shown in equation (9).
A = [20 -10;10 5];

%Substitute the value of b as shown in equation (9).
b = [100;100];

% solving using inverse
I = inv(A)*b;

%Print the result i1
fprintf('i1 = %0.2fA \n',I(1));

%Print the result i2
fprintf('i2 = %0.2fA \n',I(2));

%Print the result i3 = i1 - i2 as shown in the
%equation (6)
fprintf('i3 = %0.2fA \n',I(1)-I(2));

%Print the result i4 = i1 as shown in the
%equation (5)
fprintf('i4 = %0.2fA\n',I(1));